2021 Fall AMC 12A Problems/Problem 3

Problem

Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\frac{1}{2}$ -mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?

$\textbf{(A)}\ 2 \frac{3}{4} \qquad\textbf{(B)}\ 3 \frac{3}{4} \qquad\textbf{(C)}\ 4 \frac{1}{2} \qquad\textbf{(D)}\ 5 \frac{1}{2} \qquad\textbf{(E)}\ 6 \frac{3}{4}$

Solution 1

If Mr. Lopez chooses Route A, then he will spend $\frac{6}{30}=\frac{1}{5}$ hour, or $12$ minutes.

If Mr. Lopez chooses Route B, then he will spend $\frac{9/2}{40}+\frac{1/2}{20}=\frac{11}{80}$ hour, or $8\frac14$ minutes.

Therefore, Route B is quicker than Route A by $12-8\frac14=\boxed{\textbf{(B)}\ 3 \frac{3}{4}}$ minutes.

~MRENTHUSIASM

Solution 2

We use the equation $d=st$ to solve this problem. Recall that $1\text{ mile per hour}=\frac{1}{60}\text{ mile per minute}.$

On route $A,$ the distance is $6$ miles and the speed to travel this distance is $\frac{1}{2}$ mile per minute. Thus, the time it takes on route $A$ is $12$ minutes. For route $B$ we have to use the equation twice, once for the distance of $5- \frac{1}{2} = \frac{9}{2}$ miles with a speed of $\frac{2}{3}$ mile per minute and a distance of $\frac{1}{2}$ miles at a speed of $\frac{1}{3}$ mile per minute. Thus, the time it takes to go on Route $B$ is $\frac{9}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot 3 = \frac{27}{4} + \frac{3}{2} = \frac{33}{4}$ minutes. Thus, Route B is $12 - \frac{33}{4} = \frac{15}{4} = 3\frac{3}{4}$ faster than Route $A.$ Thus, the answer is $\boxed{\textbf{(B)}\ 3 \frac{3}{4}}.$

~NH14

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2021 Fall AMC 12A Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See Also